Recent questions in Discrete math

Discrete mathOpen question

Noor Ul Huda2022-12-20

Show that the sequence $a}_{n$ is an solution of the recurrence relation [a^n = a^n−1 + 2a^n−2 + 2n − 9 if a] if

a^n = −n + 2

Discrete mathAnswered question

Salvador Whitehead 2022-12-01

Why is $f(n)={n}^{3}$ not an onto function?

I was doing an example in a book where it asked which of these functions are one to one, the answer in the back said for $f(n)={n}^{3}$ that it is a one to one function. Then it asked which of the functions from the previous example are onto and $f(n)={n}^{3}$ was not included in the list of onto functions.

In a later example, a question asked which of these functions is a bijection, the answer included $f(x)={x}^{3}$

This is confusing because doesn't a function have to be both an onto and one to one to be a bijection? Why would the book say it was not a onto in a previous example yet declare it to be a bijection? Is the book wrong?

I was doing an example in a book where it asked which of these functions are one to one, the answer in the back said for $f(n)={n}^{3}$ that it is a one to one function. Then it asked which of the functions from the previous example are onto and $f(n)={n}^{3}$ was not included in the list of onto functions.

In a later example, a question asked which of these functions is a bijection, the answer included $f(x)={x}^{3}$

This is confusing because doesn't a function have to be both an onto and one to one to be a bijection? Why would the book say it was not a onto in a previous example yet declare it to be a bijection? Is the book wrong?

Discrete mathAnswered question

Callie Johnston 2022-11-29

What type of material allows electrons to flow freely?

Conductor

Resistor

Hestitation

Regulator

Conductor

Resistor

Hestitation

Regulator

Discrete mathAnswered question

Kirsten Bishop 2022-11-29

Showing that if n is a natural number larger than 3, then $n!>{2}^{n}$

Showing that if n is a natural number larger than $3$, then $n!>{2}^{n}$

My try:

Base Case:

If $n=4$, then $4!>{2}^{4}$

$24>16$

So, the base case is true.

Assuming $P(k)$ is true.

$k!>{2}^{k}$

Now we need to show that $P(k+1)$ is true.

$(k+1)!={2}^{k+1}$

Proof:

$(k+1)!>(k+1)k!$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(k+1){2}^{k}$

After this I have no idea how to solve further.

Can anyone explain how to continue.

Showing that if n is a natural number larger than $3$, then $n!>{2}^{n}$

My try:

Base Case:

If $n=4$, then $4!>{2}^{4}$

$24>16$

So, the base case is true.

Assuming $P(k)$ is true.

$k!>{2}^{k}$

Now we need to show that $P(k+1)$ is true.

$(k+1)!={2}^{k+1}$

Proof:

$(k+1)!>(k+1)k!$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(k+1){2}^{k}$

After this I have no idea how to solve further.

Can anyone explain how to continue.

Discrete mathAnswered question

Urijah Zhang 2022-11-27

A set X with cardinality X has how many elements in its power set (the set of all subsets of a set)?

a. $|x{|}^{2}$

b. $|X\times X|$

c. ${2}^{|x|}$

d. $|X\cup X|$

a. $|x{|}^{2}$

b. $|X\times X|$

c. ${2}^{|x|}$

d. $|X\cup X|$

Discrete mathAnswered question

neudateaLp 2022-11-22

Can a dot product of a permutation of n (1,-1) with a sequence of primes generate unique numbers?

According to Fundamental Theorem of Arithmetic any positive whole number is the product of primes. Therefore, I can create unique numbers by multiplying n primes.

If I have a list of length n generated by a finite sequence in the set 1,-1, e.g., (1,1,-1,-1,1,...), and I do a dot product with a sequence of n primes starting with 3, e.g., (3,5,7,11,13,...), do I have any guarantee that I will generate unique numbers by doing different permutations of 1s and -1s?

I know that I can't if my prime's sequence start with 2, $[1,1,-1]\cdot [2,3,5]=[-1,-1,1]\cdot [2,3,5]$, but I'm not sure for the cases where my sequence starts with higher primes.

According to Fundamental Theorem of Arithmetic any positive whole number is the product of primes. Therefore, I can create unique numbers by multiplying n primes.

If I have a list of length n generated by a finite sequence in the set 1,-1, e.g., (1,1,-1,-1,1,...), and I do a dot product with a sequence of n primes starting with 3, e.g., (3,5,7,11,13,...), do I have any guarantee that I will generate unique numbers by doing different permutations of 1s and -1s?

I know that I can't if my prime's sequence start with 2, $[1,1,-1]\cdot [2,3,5]=[-1,-1,1]\cdot [2,3,5]$, but I'm not sure for the cases where my sequence starts with higher primes.

Discrete mathAnswered question

Noe Cowan 2022-11-18

Maximize sum of paired products without replacement from {1,…,2n}

Given {1,…,2n}, how do you pair off the elements such that: when you multiply within each of the n pairs, the resulting products have maximal sum?

Here is an example: For $n=2$, this would mean pairing off elements of {1,2,3,4}, taking the product within each pair, and then adding all of those products up: where the goal is to maximize the resulting sum.

In this example, you could use:

$\{1,2\},\{3,4\}\to 1\cdot 2+3\cdot 4=2+12=14$

$\{1,3\},\{2,4\}\to 1\cdot 3+2\cdot 4=3+8=11$

$\{1,4\},\{2,3\}\to 1\cdot 4+2\cdot 3=4+6=10$

Since the pairwise product sums in this case are 14,11, and 10, we find that the maximum is 14. In particular, the maximum occurs when we pair off consecutive numbers. I suspect that the consecutive pairing may always be optimal (and that pairing first with last, second with second to last, etc may always result in the minimal pairwise product sum).

I'm not sure how to prove (or disprove!) this, and would appreciate a proof or a pointer to one. I suspect this problem has already been considered, so I include a ref-req tag as there may be terminology unfamiliar to me. If there are generalizations of this problem (e.g., starting with a set other than the first consecutive 2n positive integers) then related references will be most welcome, too!

Given {1,…,2n}, how do you pair off the elements such that: when you multiply within each of the n pairs, the resulting products have maximal sum?

Here is an example: For $n=2$, this would mean pairing off elements of {1,2,3,4}, taking the product within each pair, and then adding all of those products up: where the goal is to maximize the resulting sum.

In this example, you could use:

$\{1,2\},\{3,4\}\to 1\cdot 2+3\cdot 4=2+12=14$

$\{1,3\},\{2,4\}\to 1\cdot 3+2\cdot 4=3+8=11$

$\{1,4\},\{2,3\}\to 1\cdot 4+2\cdot 3=4+6=10$

Since the pairwise product sums in this case are 14,11, and 10, we find that the maximum is 14. In particular, the maximum occurs when we pair off consecutive numbers. I suspect that the consecutive pairing may always be optimal (and that pairing first with last, second with second to last, etc may always result in the minimal pairwise product sum).

I'm not sure how to prove (or disprove!) this, and would appreciate a proof or a pointer to one. I suspect this problem has already been considered, so I include a ref-req tag as there may be terminology unfamiliar to me. If there are generalizations of this problem (e.g., starting with a set other than the first consecutive 2n positive integers) then related references will be most welcome, too!

Discrete mathAnswered question

atgnybo4fq 2022-11-17

Counting words of length n from k-sized alphabet with no substring of k consecutive distinct letters

How many words of length n are there, if we have an alphabet of k distinct letters, but the words cannot contain any substring that is made of k consecutive distinct letters, i.e, no k-length substring that consists of the entire alphabet?

Other than that, no restrictions apply: any amount of distinct letters may be used throughout the entire word, and any letter can be used as many times as we like, as long as every substring inside the length n word complies with the rules above.

There is the obvious case of $k=2$ which results in 2 words, for every n, because you can only start with either letter, and they alternate.

For the larger case, I have come up with a recursive formula:

$C(n,k)=f(0,0,k)$

$f(i,d,k)=\{\begin{array}{ll}{\displaystyle (k-d)\cdot f(i+1,d+1,k)+\sum _{c=1}^{d}f(i+1,c,k)}& i<n,d<k\\ 1& i=n,d<k\\ 0& i>n\\ 0& d\ge k\end{array}$

With d being the current length of consecutive distinct letters, and i the current word length.

At every step, we can either use a letter other than the previous d letters, in which case the length is increased by one, and the chain-length d of distinct consecutive letters is also increased by one. In this case, there are $(k-d)$ such letters we can use at the stage, each subsequently resulting in the same contribution.

Or, we can use a letter already in the last d letters. In this case, the position of the letter matters. If we use the last of the d letters, a whole new chain begins. If instead we use the second last d letter, then a chain of length 2 of consecutive distinct letters begins. The 3rd last would result in a chain of length 3 and so on.

I was wondering if there is a another way to count the amount of such words, perhaps using matrices or combinatorics?

How many words of length n are there, if we have an alphabet of k distinct letters, but the words cannot contain any substring that is made of k consecutive distinct letters, i.e, no k-length substring that consists of the entire alphabet?

Other than that, no restrictions apply: any amount of distinct letters may be used throughout the entire word, and any letter can be used as many times as we like, as long as every substring inside the length n word complies with the rules above.

There is the obvious case of $k=2$ which results in 2 words, for every n, because you can only start with either letter, and they alternate.

For the larger case, I have come up with a recursive formula:

$C(n,k)=f(0,0,k)$

$f(i,d,k)=\{\begin{array}{ll}{\displaystyle (k-d)\cdot f(i+1,d+1,k)+\sum _{c=1}^{d}f(i+1,c,k)}& i<n,d<k\\ 1& i=n,d<k\\ 0& i>n\\ 0& d\ge k\end{array}$

With d being the current length of consecutive distinct letters, and i the current word length.

At every step, we can either use a letter other than the previous d letters, in which case the length is increased by one, and the chain-length d of distinct consecutive letters is also increased by one. In this case, there are $(k-d)$ such letters we can use at the stage, each subsequently resulting in the same contribution.

Or, we can use a letter already in the last d letters. In this case, the position of the letter matters. If we use the last of the d letters, a whole new chain begins. If instead we use the second last d letter, then a chain of length 2 of consecutive distinct letters begins. The 3rd last would result in a chain of length 3 and so on.

I was wondering if there is a another way to count the amount of such words, perhaps using matrices or combinatorics?

Discrete mathAnswered question

mxty42ued 2022-11-17

Maximum number of edges in a graph satisfying conditions

There is a graph with 40 vertices. It is known that any edge has at least one endpoint, on which no more than four other edges are incident. What is the maximum number of edges that this graph can have? No multiple edges or loops are allowed.

There is a graph with 40 vertices. It is known that any edge has at least one endpoint, on which no more than four other edges are incident. What is the maximum number of edges that this graph can have? No multiple edges or loops are allowed.

Discrete mathAnswered question

MISA6zh 2022-11-17

Compute summation of modules expression?

In particular, what I want to look at is the sum

$\sum _{k=1}^{n}(pk\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}q))$

where $p,q\in {\mathbb{Z}}_{\ge 1}$ can be assumed to be coprime but it would be best if solved in the fullest generality. In the above expression, n is a variable, p,q are fixed, and a(modb) means taking the representative set $\{0,1,2,...,b-1\}$. For example, $7\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}3)=1$ is the only value we agree upon and $7\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}3)\ne -2$.

The problem with this is that the list of representatives are permuted by p and hence the methods presented in the initial link are no longer valid.

It would be nice if we can come up with a closed form, but a really tight upper bound of the expression also works.

In particular, what I want to look at is the sum

$\sum _{k=1}^{n}(pk\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}q))$

where $p,q\in {\mathbb{Z}}_{\ge 1}$ can be assumed to be coprime but it would be best if solved in the fullest generality. In the above expression, n is a variable, p,q are fixed, and a(modb) means taking the representative set $\{0,1,2,...,b-1\}$. For example, $7\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}3)=1$ is the only value we agree upon and $7\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}3)\ne -2$.

The problem with this is that the list of representatives are permuted by p and hence the methods presented in the initial link are no longer valid.

It would be nice if we can come up with a closed form, but a really tight upper bound of the expression also works.

Discrete mathAnswered question

Yaretzi Mcconnell 2022-11-15

For each value for a and b, being real numbers, ${a}^{2}+{b}^{2}\ge ab$

Should I solve this by replacing all possible real number formulas in these to be able to prove this? As an example for odd numbers, by replacing $2x+1$, etc...?

Should I solve this by replacing all possible real number formulas in these to be able to prove this? As an example for odd numbers, by replacing $2x+1$, etc...?

Discrete mathAnswered question

Nico Patterson 2022-11-11

Representative Set for Relation S on $\mathbb{N}\to \{0,1\}$ s.t. $\u27e8f,g\u27e9\in S\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{\exists}$ bijection $h:\mathbb{N}\to \mathbb{N}$ s.t.$f=g\circ h$

Problem: Define a relation S on $\mathbb{N}\to \{0,1\}$ as follows:$\u27e8f,g\u27e9\in S\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$ there exists a bijection $h:\mathbb{N}\to \mathbb{N}$ s.t. $f=g\circ h$.

S is an equivalence relation on $\mathbb{N}\to \{0,1\}$ (no need to prove this). Write a Representative set for the relation S. There's no need to prove that the relation you wrote is indeed a Representative set.

Reminder: Suppose $T\subseteq X\times X$ is an equivalence relation over X. $\text{}\text{}A\subseteq X$ will be called a Representative set of T, if it occurs that: $\mathrm{\forall}x\in X.|[x{]}_{T}\cap A|=1$.

Attempt: I don't really know what Representative set to define. It seems to me I'm missing something simple here. I tried to look at the functions: ${f}_{1}(n)=0,{f}_{2}(n)=1,{f}_{3}(n)=\{\begin{array}{ll}0& \text{n=0}\\ 1& \text{else}\end{array}$, ${f}_{4}(n)=\{\begin{array}{ll}0& n\in {\mathbb{N}}_{even}\\ 1& n\in {\mathbb{N}}_{odd}\end{array},\mathrm{\forall}n\in \mathbb{N}$. None of these functions relate through relation S since there does not exist a bijection between them. I feel lost, do you have any idea what to do?

Problem: Define a relation S on $\mathbb{N}\to \{0,1\}$ as follows:$\u27e8f,g\u27e9\in S\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$ there exists a bijection $h:\mathbb{N}\to \mathbb{N}$ s.t. $f=g\circ h$.

S is an equivalence relation on $\mathbb{N}\to \{0,1\}$ (no need to prove this). Write a Representative set for the relation S. There's no need to prove that the relation you wrote is indeed a Representative set.

Reminder: Suppose $T\subseteq X\times X$ is an equivalence relation over X. $\text{}\text{}A\subseteq X$ will be called a Representative set of T, if it occurs that: $\mathrm{\forall}x\in X.|[x{]}_{T}\cap A|=1$.

Attempt: I don't really know what Representative set to define. It seems to me I'm missing something simple here. I tried to look at the functions: ${f}_{1}(n)=0,{f}_{2}(n)=1,{f}_{3}(n)=\{\begin{array}{ll}0& \text{n=0}\\ 1& \text{else}\end{array}$, ${f}_{4}(n)=\{\begin{array}{ll}0& n\in {\mathbb{N}}_{even}\\ 1& n\in {\mathbb{N}}_{odd}\end{array},\mathrm{\forall}n\in \mathbb{N}$. None of these functions relate through relation S since there does not exist a bijection between them. I feel lost, do you have any idea what to do?

Discrete mathAnswered question

Jared Lowe 2022-11-05

How to arrive at this integral?

However, I am not sure how to arrive at this:

$$\begin{array}{rl}\phantom{=}& {{\int}_{0}^{1}{x}^{n+2}(1-x{)}^{n}\phantom{\rule{thinmathspace}{0ex}}dx}+2{\int}_{0}^{1}{x}^{n+1}(1-x{)}^{n+1}\phantom{\rule{thinmathspace}{0ex}}dx+{{\int}_{0}^{1}{x}^{n}(1-x{)}^{n+2}\phantom{\rule{thinmathspace}{0ex}}dx}\\ =& 2I(n+1)+{2{\int}_{0}^{1}{x}^{n+2}(1-x{)}^{n}\phantom{\rule{thinmathspace}{0ex}}dx}\end{array}$$

and why the highlighted part is equal to each other? What type of integral is that?

However, I am not sure how to arrive at this:

$$\begin{array}{rl}\phantom{=}& {{\int}_{0}^{1}{x}^{n+2}(1-x{)}^{n}\phantom{\rule{thinmathspace}{0ex}}dx}+2{\int}_{0}^{1}{x}^{n+1}(1-x{)}^{n+1}\phantom{\rule{thinmathspace}{0ex}}dx+{{\int}_{0}^{1}{x}^{n}(1-x{)}^{n+2}\phantom{\rule{thinmathspace}{0ex}}dx}\\ =& 2I(n+1)+{2{\int}_{0}^{1}{x}^{n+2}(1-x{)}^{n}\phantom{\rule{thinmathspace}{0ex}}dx}\end{array}$$

and why the highlighted part is equal to each other? What type of integral is that?

Discrete mathAnswered question

MMDCCC50m 2022-11-05

Find a recursion formula for combinatorial problem

Let ${C}_{n}$ be the number of sequences with a length of n, which their elements belong to {0,1,2}, and they don't contain the following sequences: 11,21. Find a recursion formula with starting conditions for ${C}_{n}$.

Let ${C}_{n}$ be the number of sequences with a length of n, which their elements belong to {0,1,2}, and they don't contain the following sequences: 11,21. Find a recursion formula with starting conditions for ${C}_{n}$.

Discrete mathAnswered question

Noe Cowan 2022-11-02

What is the most general distribution for which $E[1/x]=1/E[x]$?

What is the most general distribution for which the expected value of the multiplicative inverse equals the multiplicative inverse of the expected value?

Motivation: I'm into modelling dynamics on graphs and I found a problem which is easily solvable in cases where the degree distribution of the vertices is a distribution where $E[1/k]=1/E[k]$. (${k}_{i}$ is the degree of the ith vertex) From this solution I may gain an insight into how to unify multiple models.

So particularly I'm looking for a distribution which consists of non-negative, finite integers. But I'm also interested in continuous solutions. Distributions where $E[1/{k}^{n}]=1/E[{k}^{n}]$ may also help unifying the models.

What I do know so far, that ${k}_{i}=1$ is a particular solution. In the continuous case every function where $f(x)=f(1/x)$ and $E[x]=1$ is a solution. I know what momentum generating functions are and they seem like a good direction to try in, but I failed so far.

What is the most general form of this distribution? Does it have a name? It sounds like something trivial, like a "famous" distribution, but I can't find it.

What is the most general distribution for which the expected value of the multiplicative inverse equals the multiplicative inverse of the expected value?

Motivation: I'm into modelling dynamics on graphs and I found a problem which is easily solvable in cases where the degree distribution of the vertices is a distribution where $E[1/k]=1/E[k]$. (${k}_{i}$ is the degree of the ith vertex) From this solution I may gain an insight into how to unify multiple models.

So particularly I'm looking for a distribution which consists of non-negative, finite integers. But I'm also interested in continuous solutions. Distributions where $E[1/{k}^{n}]=1/E[{k}^{n}]$ may also help unifying the models.

What I do know so far, that ${k}_{i}=1$ is a particular solution. In the continuous case every function where $f(x)=f(1/x)$ and $E[x]=1$ is a solution. I know what momentum generating functions are and they seem like a good direction to try in, but I failed so far.

What is the most general form of this distribution? Does it have a name? It sounds like something trivial, like a "famous" distribution, but I can't find it.

Discrete mathAnswered question

Joglxym 2022-11-02

Let $f:A\to B$ and $g:C\to D$. Define$$f\times g=\{((a,c),(b,d)):(a,b)\in f\text{and}(c,d)\in g\}.$$Prove that $f\times g$ is a function from $A\times C$ to $B\times D$.

My proof (so far):

Let $(a,c)\in A\times C$, then $a\in A$ and $c\in C$.

Let $(b,d)\in B\times D$, $b\in B$ and $d\in D$.

If $p\in A\times C$ then $p=(a,c)$.

Define $(f\times g)(p)$ as $(f\times g)(p)=(f\times g)(a,c)=(f(a),g(c))=(b,d)$, however $b\in B$ and $d\in D$ and $(b,d)\in B\times D$.

This shows $f\times g$ is a function from $A\times C$ to $B\times D$ as $(f\times g):A\times C\to B\times D$ then $(a,b)\to (f(a),g(c))$.

I know that it is not well written out, but I was wondering if my thought process so far made sense or if I accidentally missed something. Additionally, I am unsure of what the next step may be. As a sidenote, I am worried that this does not work for the given definition of $f\times g$.

My proof (so far):

Let $(a,c)\in A\times C$, then $a\in A$ and $c\in C$.

Let $(b,d)\in B\times D$, $b\in B$ and $d\in D$.

If $p\in A\times C$ then $p=(a,c)$.

Define $(f\times g)(p)$ as $(f\times g)(p)=(f\times g)(a,c)=(f(a),g(c))=(b,d)$, however $b\in B$ and $d\in D$ and $(b,d)\in B\times D$.

This shows $f\times g$ is a function from $A\times C$ to $B\times D$ as $(f\times g):A\times C\to B\times D$ then $(a,b)\to (f(a),g(c))$.

I know that it is not well written out, but I was wondering if my thought process so far made sense or if I accidentally missed something. Additionally, I am unsure of what the next step may be. As a sidenote, I am worried that this does not work for the given definition of $f\times g$.

Discrete mathAnswered question

Maribel Vang 2022-10-28

Extracting even / odd part of summation trick

Given a function f(x), we know its even part is given as $\frac{f(x)+f(-x)}{2}$ and its odd part is given by $\frac{f(x)-f(-x)}{2}$.

Consider a discrete sequence given by ${a}_{j}$ for $j\ge 1$, then the sum of the terms in the sequence till n terms is given as

$$S=\sum _{j}^{n}{a}_{j}$$

Suppose I wanted to get the sum of the even terms in the above expression; then

$${S}_{odd}=\sum _{j}^{n}\frac{{a}_{-j}+{a}_{j}}{2}$$

and, for odd,

$${S}_{odd}=\sum _{j}^{n}\frac{{a}_{j}-{a}_{-j}}{2}$$

But wait, our sequence was defined for $j\ge 1$. Well, here's the thing: ${a}_{j}$ is some function of j, extending the domain to negative integers and evaluating the function gives the right answer... but I can't understand why the continuous function trick extended to here.

Examples:

Sum of first n numbers given as $\frac{n(n+1)}{2}$, sum of first n odds will be given as: $\frac{n(n+1)}{2}-\frac{n(1-n)}{2}$

My attempt at finding an exact connection: To the discrete sequence $\frac{n(n+1)}{2}$, we can associate a function $f(x)=\frac{x(x+1)}{2}$ and we can think of the summation as summing this function at several different input points i.e:

$$S=\sum _{j}^{n}{a}_{j}\to \sum _{k}^{n}f(x+k)$$

Then we apply the even odd decomposition and return back to the sequence world.

My question: Does there exist an association for every sequence with a function? If not, what is the criterion for an association to exist?

Given a function f(x), we know its even part is given as $\frac{f(x)+f(-x)}{2}$ and its odd part is given by $\frac{f(x)-f(-x)}{2}$.

Consider a discrete sequence given by ${a}_{j}$ for $j\ge 1$, then the sum of the terms in the sequence till n terms is given as

$$S=\sum _{j}^{n}{a}_{j}$$

Suppose I wanted to get the sum of the even terms in the above expression; then

$${S}_{odd}=\sum _{j}^{n}\frac{{a}_{-j}+{a}_{j}}{2}$$

and, for odd,

$${S}_{odd}=\sum _{j}^{n}\frac{{a}_{j}-{a}_{-j}}{2}$$

But wait, our sequence was defined for $j\ge 1$. Well, here's the thing: ${a}_{j}$ is some function of j, extending the domain to negative integers and evaluating the function gives the right answer... but I can't understand why the continuous function trick extended to here.

Examples:

Sum of first n numbers given as $\frac{n(n+1)}{2}$, sum of first n odds will be given as: $\frac{n(n+1)}{2}-\frac{n(1-n)}{2}$

My attempt at finding an exact connection: To the discrete sequence $\frac{n(n+1)}{2}$, we can associate a function $f(x)=\frac{x(x+1)}{2}$ and we can think of the summation as summing this function at several different input points i.e:

$$S=\sum _{j}^{n}{a}_{j}\to \sum _{k}^{n}f(x+k)$$

Then we apply the even odd decomposition and return back to the sequence world.

My question: Does there exist an association for every sequence with a function? If not, what is the criterion for an association to exist?

Dealing with discrete Math is an interesting subject because discrete Math equations can be encountered basically anywhere from scheduling of sports games and live shows to education where each person is examined online. It is a reason why discrete math questions that we have collected for you are aimed at solutions that go beyond equations to provide you with the answers that will help you understand the concept. Still, discrete Math equations are explained as well by turning to problems in computer science, programming, software, and cryptography among other interesting subjects like software and mobile apps development.